Question

An Olympic-class sprinter starts a race with an acceleration of\({\bf{4}}{\bf{.50}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\). (a) What is her speed 2.40 s later? (b) Sketch a graph of her position vs. time for this period.

Short answer

(a) The speed of sprinter is\({\bf{10}}{\bf{.8}}\;{\bf{m/s}}\).

(b) The sketch of position and time for sprinter is given in Figure (1).

Step-by-step solution

Determination of speed of sprinter

Given Data:

The acceleration of sprinter is\(a = 4.50\;{\rm{m}}/{{\rm{s}}^2}\)

The initial speed of car is\(u = 0\;{\rm{m}}/{\rm{s}}\)

The duration for speed of sprinter is\(t = 2.40\;{\rm{s}}\)

The speed of sprinter is found by using the first equation of motion as the acceleration of sprinter is uniform.

The speed of sprinter is given as

\(v = u + at\)

Here,\(v\)is the speed of sprinter.

Substitute all the values in the above equation.

\(\begin{array}{c}v = 0\;{\rm{m}}/{\rm{s}} + \left( {4.50\;{\rm{m}}/{{\rm{s}}^2}} \right)\left( {2.40\;{\rm{s}}} \right)\\v = 10.8\;{\rm{m}}/{\rm{s}}\end{array}\)

Therefore, the speed of sprinter is \(10.8\;{\rm{m}}/{\rm{s}}\).

Determination of sketch of position and time for sprinter

The position of sprinter is given as

\(X = ut + \frac{1}{2}a{t^2}\)

Here,\(X\)is the position of sprinter at time\(t\).

Substitute all the values in the above equation.

\[\begin{array}{l}X = \left( 0 \right)t + \frac{1}{2}\left( {4.50\;{\rm{m}}/{{\rm{s}}^2}} \right){t^2}\\X = 2.25{t^2}\end{array}\]

Sketch the above equation for the position and time graph of the sprinter.

It is as given in Figure (1).